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Q.

A and B are two points separated by a distance 5cm. Two charges 10 μC and 20 μC are placed at A and B. The resultant electric intensity at a point P outside the charges at a distance 5cm from 10 μC is

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a

zero

b

9 × 106 N/C away from 10 μC

c

56 × 106 N/C towards 10 μC

d

54 × 106 N/Caway from 10 μC

answer is A.

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Detailed Solution

from principle of superposition

Ep=EA+EB

Ep=9×109qArA2+qBrB2

Ep=54×106N/C( away from 10μC)

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